The escape velocity of a body from the earth is `11.2 km//s`. If a body is projected with a velocity twice its escape velocity, then the velocity of the body at infinity is (in `km//s`)

Options

A. `19.4`
B. `194`
C. `1.94`
D. `0.194`

Explanation

To solve the problem, we need to determine the velocity of a body projected with a velocity twice the escape velocity of Earth when it reaches infinity. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify Escape Velocity**:
The escape velocity \( v_e \) from the surface of the Earth is given as \( 11.2 \, \text{km/s} \).

2. **Calculate Initial Velocity**:
The body is projected with a velocity \( v_1 \) which is twice the escape velocity:
\[ v_1 = 2 \times v_e = 2 \times 11.2 \, \text{km/s} = 22.4 \, \text{km/s} \]

3. **Use Conservation of Energy**:
According to the conservation of energy, the total mechanical energy at the surface of the Earth must equal the total mechanical energy at infinity. The total energy \( E \) can be expressed as:
\[ E = \text{Kinetic Energy} + \text{Potential Energy} \]

4. **Calculate Kinetic and Potential Energy at the Surface**:
- Kinetic Energy at the surface:
\[ KE_{\text{surface}} = \frac{1}{2} m v_1^2 = \frac{1}{2} m (22.4)^2 \]
- Potential Energy at the surface:
\[ PE_{\text{surface}} = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

5. **Total Energy at the Surface**:
The total energy at the surface is:
\[ E_{\text{surface}} = KE_{\text{surface}} + PE_{\text{surface}} = \frac{1}{2} m (22.4)^2 - \frac{GMm}{R} \]

6. **Calculate Kinetic Energy at Infinity**:
At infinity, the potential energy is zero, and the total energy is purely kinetic:
\[ E_{\text{infinity}} = KE_{\text{infinity}} = \frac{1}{2} m v_2^2 \]

7. **Set Total Energies Equal**:
Since total energy is conserved:
\[ \frac{1}{2} m (22.4)^2 - \frac{GMm}{R} = \frac{1}{2} m v_2^2 \]

8. **Simplify and Solve for \( v_2 \)**:
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} (22.4)^2 - \frac{GM}{R} = \frac{1}{2} v_2^2 \]
Rearranging gives:
\[ v_2^2 = (22.4)^2 - \frac{2GM}{R} \]
We know that \( \frac{GM}{R} = \frac{1}{2} v_e^2 \), thus:
\[ v_2^2 = (22.4)^2 - (11.2)^2 \]
\[ v_2^2 = 22.4^2 - 11.2^2 = 4 \times 11.2^2 = 4 \times 125.44 = 501.76 \]
\[ v_2 = \sqrt{501.76} \approx 22.4 \, \text{km/s} \]

9. **Final Calculation**:
To find the final velocity at infinity:
\[ v_2 = \sqrt{6} \cdot v_e \approx 1.732 \cdot 11.2 \approx 19.39 \, \text{km/s} \]

### Final Answer:
The velocity of the body at infinity is approximately \( 19.4 \, \text{km/s} \).

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